Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL−1?
Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO3) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol−1
Then, number of moles of $\mathrm{HNO}_{3}=\frac{68}{63} \mathrm{~mol}$
$=1.079 \mathrm{~mol}$
Given,
Density of solution = 1.504 g mL−1
$\therefore$ Volume of $100 \mathrm{~g}$ solution $=\frac{100}{1.504} \mathrm{~mL}$
$=66.49 \mathrm{~mL}$
$=66.49 \times 10^{-3} \mathrm{~L}$
Molarity of solution $=\frac{1.079 \mathrm{~mol}}{66.49 \times 10^{-3} \mathrm{~L}}$
$=16.23 \mathrm{M}$